# Solution to the Two Squares on a Hexagon Puzzle +-- {.image} [[TwoSquaresonaHexagon.png:pic]] > Two squares and a regular hexagon. What’s the angle? =-- ## Solution by Angle Properties +-- {.image} [[TwoSquaresonaHexagonLabelled.png:pic]] =-- With the points labelled as above, let $x = B \hat{O} F$. Then as angle $O \hat{B} F$ is a right angle and [[angles in a triangle]] add up to $180^\circ$, $B \hat{F} O = 90^\circ - x$. Since $K$ is the [[midpoint]] of $O F$ and triangle $O B F$ is a [[right-angled triangle]], triangle $B K F$ is [[isosceles]] and so angle $K \hat{B} F = K \hat{F} B = 90^\circ - x$. This means that angle $B \hat{K} F = 180^\circ - 2 \times (90^\circ - x) = 2 x$ and so angle $B \hat{K} H = 90^\circ + 2 x$. The length of $B K$ is equal to that of $K F$ which is equal to the length of $K H$. Hence triangle $B K H$ is also [[isosceles]] and so angle $K \hat{B} H$ is one half of $180^\circ - (90^\circ + 2 x)$ which is $45^\circ - x$. Then: $$ H \hat{B} F = K \hat{B} F - K \hat{B} H = (90^\circ - x ) - (45^\circ - x) = 45^\circ) $$ To get the requested angle, note that $E \hat{J} I = H \hat{B} C $ by [[alternate angles]] and $D \hat{E} J = 60^\circ$ as the [[exterior angle]] of a [[regular hexagon]]. So using [[angles in a triangle]] and [[angles on a straight line]], angle $D \hat{I} J = 180^\circ - (180^\circ - 45^\circ - 60^\circ) = 105^\circ$. ## Solution by [[Circle Theorems]] Construct a circle centred on $K$ and passing through $O$. As $K H$ and $K O$ are sides of a square, $H$ is also on this circle. As $K$ is the midpoint of $O F$, $F$ is also on this circle. Finally, as triangle $O B F$ is a [[right-angled triangle]] and $O F$ is a diagonal, $B$ is also on this circle. +-- {.image} [[TwoSquaresonaHexagonCircled.png:pic]] =-- Using [[angles in the same segment are equal]], angles $F \hat{B} H$ and $F \hat{O} H$ are equal. Since $F \hat{O} H$ is the angle between a side and a diagonal of a square, it is $45^\circ$. Hence $F \hat{B} H = 45^\circ$. The requested angle then is $105^\circ$ as above. ## Solution by [[Transformation]] With the points labelled as above, the point $F$ moves on a straight line through $B$ and $C$. The point $H$ is obtained from $F$ by a fixed transformation of a rotation of $45^\circ$ anticlockwise about $O$ and scaling by $\frac{1}{\sqrt{2}}$. It therefore also moves on a straight line. To see where this straight line is, consider two special cases. In the first, $F$ is at $B$. In this configuration, $H$ is at a point so that angle $G \hat{B} H = 45^\circ$. +-- {.image} [[TwoSquaresonaHexagonFirstSpecial.png:pic]] =-- In the second, $F$ is taken to the left so that the length of $F B$ is the same as that of $O B$. This brings $H$ to coincide with $B$ and shows that $B$ is on the line that $H$ moves along. +-- {.image} [[TwoSquaresonaHexagonSecondSpecial.png:pic]] =-- Taken together, these show that $H$ moves on a line that goes through $B$ at an angle of $45^\circ$ to the horizontal. ## Solution by [[Invariance Principle]] In the configuration where $F$ is at $B$, the line $B H$ is the diagonal of a square so the angle $G \hat{H} B$ is $45^\circ$. From here, the given angle can be calculated as $105^\circ$ as above.